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Ber [7]
2 years ago
6

The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equa

tion for N to derive this relationship.
Hint: when t = t(1/2), N/No = 0.5 (1 point)

A sample of carbon-14 initially consists of 5 × 1024 particles. Carbon-14 has a half-life of 5730 years.

a. What is the decay constant for carbon-14? (Answer in units of yr–1.) (1 point)
b. How many radioactive particles of the sample remain after 100 years? (1 point)
c. What percentage of the radioactive particles remains after 500 years? (1 point)
d. How many radioactive particles of the sample remain after 1000 years? (1 point)
e. How much time will it take for 50% of the particles to decay? (1 point)
f. How much time will it take for 99% of the particles to decay? (1 point)
g. How many half-lives will it take for 99% of the sample to decay? (1 point)
h. What is the initial decay rate of the sample? (Answer in decays/yr.) (1 point)
i. After 200 years, what is the decay rate of the sample? (1 point)
j. How long will it take for the decay rate to decrease to 1015 decays/year? (1 point)
k. How many half-lives have passed after the time you found in part (j)? (1 point)
Physics
1 answer:
pshichka [43]2 years ago
6 0

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda

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What is your theory on frame of reference?
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What is your theory on frame of reference?

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8 0
3 years ago
When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

8 0
3 years ago
As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
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